# Weak convergence and bounded sequence on a Riemannian manifold

Consider a Riemannian manifold $\mathcal{M}$ modeled on a (possibly infinitely dimensional) Hilbert space. Suppose that $\{p_i\} \subset \mathcal{M}$ and $p_i \to p$ .

We say that a sequence of covectors $v^*_i \in T^*_{p_i}\mathcal{M}$ is weakly converges to $v \in T_p^*\mathcal{M}$ if for every smooth function $f$ on $\mathcal{M}$ we have $$\langle v^*_i, df(p_i) \rangle \to \langle v^*, df(p) \rangle.$$

Question: Is it true that if $\{v_i^*\}$ is bounded then it has a weakly convergent subsequence?

• When I corrected the title, I missed the obvious grammatical error in the very first word! [hangs head in shame] – David Roberts Aug 6 '17 at 11:24

## 1 Answer

Sorry, I did not read it carefully enough. Here is the revised version.

The answer is yes, if it is a strong Riemannian manifold modelled on Hilbert space (strong: generated the topology on each tangent space).

Namely, since $p_i \to p$ we may assume without loss that all $p_i$ are in one chart, so we are in a Hilbert space $H$, and the Riemannian metric is bounded by a contant times the standard one on an open subset containing all $p_i$.

By assumption, the $v_i^*$ are bounded, thus contained in a closed ball which is compact in the weak topology, thus there is a weakly convergent subsequence. Note that for each $(p,\alpha)\in T^*_p H$ there exists a smooth functions $f$ such that $df(p) = \alpha$. So we have the the weak topology to deal with.